# Null Quantification with empty domain

Is Null Quantification false?

Null Quantification state that $$\exists x [P \lor Q(x)]\equiv P\lor\exists x[Q(x)]$$

; $$P$$ represents any wff in which $$x$$ does not occur free.

If we set domain of discourse be empty and let T be tautology

we will get $$\exists x [\text{T} \lor Q(x)]$$ is false. But $$\text{T}\lor\exists x[Q(x)]$$ is true.

• We do not normally consider empty domains. There are many inference rules which also don't work for empty domains. Commented Oct 23, 2018 at 11:04

Long comment

Not all valid sentences are inclusive valid, i.e. true in all interpretations, including the interpretation with an empty domain.

Example: $$(\forall x) \bot$$ is true in the empty domain (because all universaly quantified formulas are) and thus $$(\forall x) \bot \to \bot$$ is false.

But it is valid, because all instances of $$(\forall x) \varphi \to \varphi$$ are.

If so, it seems that you have proved that :

$$∃x[ \top \lor Q(x)] \leftrightarrow (\top \lor ∃x[Q(x)])$$

is valid but not inclusive valid.

See e.g. E.Mendelson, Introduction to Mathematical Logic (6th ed, 2015) Ch.2.16 Quantification Theory Allowing Empty Domains, page 146-on.